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Phase 1/2 Designs With Efficacy and Toxicity · Tutorial 617 of 799

Deriving Joint Response Probabilities Under Independence

Learn how dose-specific efficacy and toxicity probabilities determine a complete joint outcome distribution under independence, and how to recognize when that simplification is questionable.

Advanced 10 min read

What You'll Learn

  • Define dose-specific independence for binary efficacy and toxicity outcomes.
  • Derive the four joint probabilities from the two marginal probabilities.
  • Check that the resulting joint table has valid probabilities and recovers its marginals.
  • Apply the calculation to hypothetical dose-specific probabilities and observed outcome counts.
  • Distinguish within-dose independence from dependence that can appear after combining dose groups.
  • Identify clinical and data features that make independence implausible.

Why Derive the Joint Distribution?

In a dose-finding trial, efficacy and toxicity are distinct binary outcomes for each patient. A dose-specific efficacy probability and a dose-specific toxicity probability describe each outcome separately, but decisions that consider benefit and harm often require the probabilities of their four possible combinations. The independence assumption supplies a simple, explicit way to calculate those joint probabilities from the two marginal probabilities.

That simplicity comes at a cost: independence is a substantive assumption about patients’ outcomes, not an arithmetic convenience. It should be made clear, checked against clinical knowledge and available data, and not mistaken for a fact implied by fitting separate marginal models.

Definition: At dose \(d\), let \(E=1\) indicate that a patient experiences the defined efficacy outcome within its specified assessment window, and let \(T=1\) indicate that the patient experiences the defined toxicity outcome within its specified assessment window. Let \(E=0\) and \(T=0\) indicate that the respective outcomes do not occur. Write \(p_E(d)=P(E=1\mid D=d)\) and \(p_T(d)=P(T=1\mid D=d)\), where \(D\) is the dose administered. Efficacy and toxicity are independent at dose \(d\) if, for every \(e,t\in\{0,1\}\), \(P(E=e,T=t\mid D=d)=P(E=e\mid D=d)P(T=t\mid D=d)\).

The conditioning on dose matters. This definition concerns the joint outcomes of a patient at a particular dose, in the target population and under the stated outcome definitions and observation windows. It does not say that outcomes are independent of dose, that different patients’ outcomes are independent, or that efficacy and toxicity remain independent after patients from different dose groups are combined.

Deriving the Four Probabilities

At a fixed dose, abbreviate \(p_E(d)\) as \(p_E\) and \(p_T(d)\) as \(p_T\). Under independence, multiply the appropriate marginal probability for efficacy by the appropriate marginal probability for toxicity. For an outcome not to occur, use the complement of its marginal probability.

$$ \begin{aligned} P(E=1,T=1\mid D=d) &= p_Ep_T,\\ P(E=1,T=0\mid D=d) &= p_E(1-p_T),\\ P(E=0,T=1\mid D=d) &= (1-p_E)p_T,\\ P(E=0,T=0\mid D=d) &= (1-p_E)(1-p_T). \end{aligned} $$

The order in each pair is efficacy first, toxicity second. Thus, \(P(E=1,T=0\mid D=d)\) is the probability of efficacy without toxicity; it is not the probability of toxicity without efficacy. Keeping the outcome order fixed prevents a common source of table-labeling errors.

Justification: By the independence definition, the probability of each joint event is the product of its two marginal event probabilities. Since \(P(E=0\mid D=d)=1-p_E\) and \(P(T=0\mid D=d)=1-p_T\), substituting the relevant success or complement probability for each outcome gives the four expressions above.

The formulas produce a valid joint distribution whenever \(0\leq p_E\leq1\) and \(0\leq p_T\leq1\). Each cell is nonnegative because it is a product of numbers between zero and one. Their sum is

$$ p_Ep_T+p_E(1-p_T)+(1-p_E)p_T+(1-p_E)(1-p_T) =[p_E+(1-p_E)][p_T+(1-p_T)]=1. $$

They also recover the input marginals. Adding the two cells where efficacy occurs gives \(p_Ep_T+p_E(1-p_T)=p_E\). Adding the two cells where toxicity occurs gives \(p_Ep_T+(1-p_E)p_T=p_T\). These checks are useful when constructing or programming a joint probability table.

Worked Example: Calculating the Four Cells at One Dose

Suppose, hypothetically, that at a particular dose the estimated probability of efficacy is \(p_E=0.60\), and the estimated probability of toxicity is \(p_T=0.25\). Assume for this calculation that efficacy and toxicity are independent at that dose.

Both outcomes. The probability of efficacy and toxicity is \(0.60\times0.25=0.150\), or 15.0%.

Efficacy without toxicity. The probability is \(0.60\times(1-0.25)=0.60\times0.75=0.450\), or 45.0%.

Toxicity without efficacy. The probability is \((1-0.60)\times0.25=0.40\times0.25=0.100\), or 10.0%.

Neither outcome. The probability is \((1-0.60)\times(1-0.25)=0.40\times0.75=0.300\), or 30.0%.

The cells sum to \(0.150+0.450+0.100+0.300=1.000\). The efficacy marginal is \(0.150+0.450=0.600\), and the toxicity marginal is \(0.150+0.100=0.250\), so both input probabilities are recovered. Under the assumed model, 15.0% of patients experience both outcomes and 45.0% experience efficacy without toxicity.

Using the Calculation Across Doses

If marginal models provide dose-specific values \(p_E(d)\) and \(p_T(d)\), apply the same four formulas separately at each dose. This creates a complete joint distribution for each dose without changing the marginal predictions. The assumption must still be understood as applying within each dose; it does not follow merely because the efficacy and toxicity models were fitted separately.

Worked Example: Constructing Joint Probabilities at Two Doses

Consider hypothetical model predictions at two doses. At dose \(d_1\), let \(p_E(d_1)=0.35\) and \(p_T(d_1)=0.10\). At dose \(d_2\), let \(p_E(d_2)=0.70\) and \(p_T(d_2)=0.30\). For illustration, assume independence within each dose.

At \(d_1\), the probability of both outcomes is \(0.35\times0.10=0.035\). Efficacy without toxicity has probability \(0.35\times0.90=0.315\); toxicity without efficacy has probability \(0.65\times0.10=0.065\); and neither has probability \(0.65\times0.90=0.585\). The four values sum to \(1.000\). They give efficacy probability \(0.035+0.315=0.350\) and toxicity probability \(0.035+0.065=0.100\).

At \(d_2\), the probability of both outcomes is \(0.70\times0.30=0.210\). Efficacy without toxicity has probability \(0.70\times0.70=0.490\); toxicity without efficacy has probability \(0.30\times0.30=0.090\); and neither has probability \(0.30\times0.70=0.210\). These sum to \(1.000\), with marginals \(0.210+0.490=0.700\) for efficacy and \(0.210+0.090=0.300\) for toxicity.

The interpretation is dose-specific: under the independence assumption, the model assigns a 21.0% probability to both efficacy and toxicity at \(d_2\), compared with 3.50% at \(d_1\). These calculations describe hypothetical model probabilities, not observed trial frequencies or causal effects of dose.

What Independence Implies—and What It Does Not

Under independence, the probability of both outcomes is fixed at the product of the marginal probabilities. Consequently, once the marginals are specified, all four joint cells are determined. This is a strong restriction: knowing only the two marginal probabilities does not, in general, determine how often efficacy and toxicity occur together. The introductory material on dependence between efficacy and toxicity, copula models, and shared latent-variable models provides other ways to represent that relationship.

The assumption is especially consequential when the joint cells are used to assess a benefit-harm pattern. The marginal efficacy probability can be correct while the predicted frequency of patients with both efficacy and toxicity is not. A joint table should therefore be presented as a consequence of the assumption and the chosen marginals, not as an independently verified clinical fact.

Independence also does not necessarily persist when dose groups are pooled. Suppose the dose varies across patients, with dose-specific probabilities \(p_E(d)\) and \(p_T(d)\), and independence holds within each dose. If the dose distribution assigns weight \(w_d\) to dose \(d\), the pooled probability of both outcomes is

$$ P(E=1,T=1)=\sum_d w_d\,p_E(d)p_T(d). $$

The pooled marginal probabilities are \(\sum_d w_d p_E(d)\) and \(\sum_d w_d p_T(d)\). Their product need not equal the pooled probability of both outcomes. For example, if doses with higher efficacy probabilities also have higher toxicity probabilities, patients pooled across doses can show positive association even when efficacy and toxicity are independent within every dose. This is one reason to state the dose conditioning explicitly.

Worked Example: Comparing Observed Counts With Independence

In a hypothetical group of 100 patients treated at the same dose, suppose the observed counts are 30 with both efficacy and toxicity, 30 with efficacy but no toxicity, 10 with toxicity but no efficacy, and 30 with neither outcome.

The observed efficacy proportion is \((30+30)/100=0.60\), and the observed toxicity proportion is \((30+10)/100=0.40\). If independence were imposed using these marginal proportions, the predicted joint proportions would be \(0.60\times0.40=0.24\) for both outcomes, \(0.60\times0.60=0.36\) for efficacy without toxicity, \(0.40\times0.40=0.16\) for toxicity without efficacy, and \(0.40\times0.60=0.24\) for neither.

Multiplying by 100 gives predicted counts of 24, 36, 16, and 24, respectively. The observed count with both outcomes is 30, six more than the independence-based count of 24. This descriptive discrepancy suggests that independence may not represent these data well. It is not, by itself, a formal test or a conclusive finding: sampling variation, study design, outcome ascertainment, and the target population also matter.

Assessing Plausibility and Recognizing Failure

Before using the product formulas, confirm that the setup supports their interpretation. In particular, check that efficacy and toxicity are defined for the same patient population at the same dose, with clinically meaningful and clearly specified assessment windows. The outcome coding must be genuinely binary for the calculation: each patient must be classifiable as having or not having each defined outcome. If one outcome is unobserved for some patients, treating it as absent can distort the joint table.

Clinical knowledge can challenge independence even when data are sparse. A treatment response might increase exposure to a biological mechanism that also causes toxicity; patient characteristics might influence both outcomes; or early toxicity might prevent a patient from receiving enough treatment to achieve efficacy. Conversely, treatment interruption or rescue care can affect the two outcome processes in ways that make a simple product model unsuitable. The direction of association cannot be assumed from the labels “efficacy” and “toxicity” alone.

With adequate data at a dose, compare observed joint frequencies with the four probabilities implied by the marginal estimates. Large or systematic discrepancies, particularly in the both-outcomes cell, are a warning that the assumption may be inadequate. Sparse dose groups make such comparisons noisy, so lack of an obvious discrepancy is not proof of independence. Also check that model-generated marginal probabilities stay between zero and one and that each calculated table sums to one; failure of these basic checks indicates a coding or calculation error.

When dependence is clinically important or supported by the data, a joint model that allows it may be preferable. Independence is often simpler and can be useful when the available information supports only marginal modeling, but it cannot describe every clinically plausible pattern. A dependence model adds structure and may require more data or stronger modeling choices. The suitable approach depends on the trial question, endpoint definitions, and evidence; no joint model should be treated as automatically correct.

Common Mistake

Wrong reasoning: “Efficacy and toxicity are modeled separately, so their joint probability is the product of their marginal probabilities.” Separate models specify or estimate the marginals. They do not establish independence. Correction: Multiplication is justified only when independence at the relevant dose is an explicit, defensible assumption. If that assumption is not credible, the marginal probabilities alone do not determine the four joint outcome probabilities.

Key takeaway: At a fixed dose, independence gives the four joint probabilities \(p_Ep_T\), \(p_E(1-p_T)\), \((1-p_E)p_T\), and \((1-p_E)(1-p_T)\), in the order both, efficacy only, toxicity only, and neither. The arithmetic is straightforward; deciding whether the assumption is clinically plausible is the substantive task.

Check Your Understanding

Answer the questions, then compare your reasoning with the solutions below.

  1. At a hypothetical dose, \(p_E=0.50\) and \(p_T=0.20\). Under independence, what is the probability of toxicity without efficacy?
  2. For \(p_E=0.80\) and \(p_T=0.25\), calculate the probability of neither outcome and verify the two marginal probabilities from the resulting table.
  3. Why does fitting separate marginal models for efficacy and toxicity not establish independence?
  4. If independence holds within each dose, must it hold after all dose groups are pooled? Explain briefly.

Solutions: (1) \((1-0.50)\times0.20=0.10\), or 10%. (2) Neither has probability \((1-0.80)(1-0.25)=0.20\times0.75=0.15\). The four cells are 0.20, 0.60, 0.05, and 0.15; efficacy is \(0.20+0.60=0.80\), and toxicity is \(0.20+0.05=0.25\). (3) Separate marginal models provide the probability of each outcome separately, not the probability of their co-occurrence; the product requires an additional independence assumption. (4) No. Dose-specific marginal probabilities can vary together across dose groups, creating pooled association even if outcomes are independent within each dose.

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